Calculating Local Gravity: How is g = 9.81m/s^2?

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The local gravitational field, represented as g = 9.81 m/s², is calculated using Newton's law of universal gravitation, assuming Earth is a perfect sphere. The equation involves the gravitational constant (G), the mass of the Earth (M_E), and the radius of the Earth (R_E). By rearranging the formula G(M_E * m) / R_E² = mg, one can solve for g. The specific values for G, M_E, and R_E are essential for this calculation. Understanding these parameters is crucial for accurately determining the acceleration due to gravity at Earth's surface.
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Homework Statement


I am wondering how did the local graviational field (free fall acceleration) g = 9.81m/s^2 is calculated?


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The Attempt at a Solution



 
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It is usually calculated by assuming that the Earth is a perfect sphere and then by using Newton's law of universal gravitation to find the force on an object at the surface of the Earth. Input parameters are the gravitational constant and the radius and mass of the Earth.
 
So what is the actual equation used to obtain the g (with the values)?
Just out of general interest.
 
G\frac{M_E\:m}{R_E^2}=mg

Solve for g.
 
OK thank you for your time.
:)
 

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